I worked on nested loop.

y is a 5000 by 24 matrix. i need a new 46 by 24 matrix such that row 1 of y repeats 23 times and row 2 repeats 23 times. the code is given below

p=1;
k=1;
a=ones([1,24]);
while k<3
    sub=y(k,:);
    k=k+1;
    while j<24
        a(j,:)=sub;
        j=j+1;
    end
end

the answer i expect is a 46 by 24 matrix. but i get a 23 by 24 matrix.

Answer 1

[per isakson]

Assuming that your goal is to understand loops, try this

>> [ a01, a02 ] = cssm();
>> all(a01(:)==a02(:))
ans =
     1

where

function  [ a01, a02 ] = cssm()
    y   = randi( 100, [3,24] );
    a01 = nan([46,24]);
    a02 = nan([46,24]);
    kk  = 1;
    while kk<3
        sub = y(kk,:);
        kk  = kk+1;
        jj  = 1;
        while jj<24
            a01((kk-2)*23+jj,:)=sub;
            jj = jj+1;
        end
    end
    for kk = 1 : 2
        for jj = 1 : 23
            a02((kk-1)*23+jj,:) = y(kk,:);
        end
    end
end

&nbsp

Comment: &nbsp The codes based on while-loops and for-loops, respectively, produces identical answers. In this special case for-loops are easier to work with than while-loops.

"but i get a 23 by 24 matrix". That's because in the second turn of the outer loop the counter, j, has the value 24 and the code of the inner loop will not execute.

Stalin's answer provides a more efficient solution, which takes advantage of special features of Matlab.

[seema niran]

yes sir , i got the correct matrix when i incremented j (j=j+23) in the first loop. thank you for the comment

Answer 2

[Stalin Samuel]

A_new = ones(46,24)
A = rand(5000,24);
A_new(1:23,:) = A(1,:);
A_new(24:46,:) = A(2,:) ;